\documentclass[11pt]{article}

\setlength{\textwidth}{6.5in}
\setlength{\oddsidemargin}{0in}
\setlength{\evensidemargin}{0in}

\usepackage{amsmath,amscd,amsthm,amsfonts}

\theoremstyle{definition}
\newtheorem{defn}{Definition}
\newtheorem{exer}{Exercise}
\theoremstyle{plain}
\newtheorem{thm}{Theorem}
\newtheorem{lem}{Lemma}
\newtheorem{cor}{Corollary}
\newtheorem{prop}{Proposition}

\renewcommand{\P}{\mathbb{P}^1}
\renewcommand{\k}{\kappa}
\renewcommand{\H}{\mathbb{H}}
\newcommand{\R}{\mathbb{R}}
\newcommand{\F}{\mathfrak{F}}
\newcommand{\C}{\mathbb{C}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\Q}{\mathbb{Q}}
\newcommand{\N}{\mathbb{N}}
\newcommand{\G}{\Gamma}
\newcommand{\GO}[1]{\Gamma_0(#1)}
\newcommand{\res}[2]{\mathrm{Res}_{#2}(#1)}
\newcommand{\Gb}[1]{\bar{\Gamma}(#1)}
\newcommand{\SL}[1]{\mathbf{SL}_2(#1)}
\newcommand{\PSL}[1]{\mathbf{PSL}_2(#1)}
\newcommand{\GL}[1]{\mathbf{GL}_2(#1)}
\newcommand{\V}[2]{\bigl[\begin{smallmatrix}#1 \\#2\end{smallmatrix}\bigr]}
\newcommand{\z}{\zeta}
\renewcommand{\a}{\alpha_p}
\renewcommand{\b}{\beta_p}
\newcommand{\ba}{\bar{\alpha}_p}
\newcommand{\bb}{\bar{\beta}_p}
\newcommand{\s}{\sigma}
\newcommand{\e}{\epsilon}
\newcommand{\T}{\Theta}
\DeclareMathOperator{\deck}{Deck}
\DeclareMathOperator{\aut}{Aut}
\DeclareMathOperator{\id}{id}
\DeclareMathOperator{\gal}{Gal}
\DeclareMathOperator{\ord}{ord}

\title{Modular Forms, PS 7}
\author{Bryden Cais}
\begin{document}
\maketitle

\begin{exer}
    Recall we showed in class that
    \begin{align*}
        \T(s)E(z,s)=\T(s)y^s+\T(s-1/2)y^{1-s}+\sum_{r\neq 0}\frac{W_s(rz)}{\sqrt{|r|}}\left(\sum_{ab=|r|}\left(\frac{a}{b}\right)^{s-1/2}\right),
    \end{align*}
    where
    $$\T(s)=\pi^{-s}\G(s)\z(2s).$$
    Observe that $\T(s)$ has simple poles corresponding to a pole of the Gamma function at $s=0$ and a pole of $\z(2s)$ as $s=1/2$.  Since
    $\z(2s)$ has simple zeroes at $s=-n$ for any positive integer $n$, we see that $\T(s)$ is holomorphic everywhere except at $s=0,1/2$.
    The residue at $s=1/2$ is readily seen to be
    $$R_{1/2}(\T(s))=\pi^{-1/2}\G(1/2)\frac{1}{2}=\frac{1}{2},$$
    while at $0$ we have
    $$R_{0}(\T(s))=\z(0)=-\frac{1}{2}.$$
    We have seen that $W_s(z)$ is a holomorphic function of $s$, for any value of $z$, and that as $\Im(z)$ tends to infinity,
    $W_s(z)$ decays exponentially.  It follows that any pole of $\T(s)E(z,s)$ comes from the term $\T(s)y^s+\T(s-1/2)y^{1-s}$.
    Clearly this term has (potential) simple poles at $s=1/2,0,1$ and nowhere else.  By our residue calculations above, the
    residue of $\T(s)y^s+\T(s-1/2)y^{1-s}$ at $s=1/2$ is
    $$\frac{1}{2}y^{1/2}-\frac{1}{2}y^{1-1/2}=0,$$ so that, in fact, $E(z,s)$ is holomorphic at $s=1/2$.  A short computation shows that
    the residue at $0$ is
    $$-\frac{1}{2},$$ while at $1$ it is
    $$R_{1}(\T(s-1/2))=\frac{1}{2}.$$  Since $\T(s)E(z,s)$ has a simple pole of residue $-\frac{1}{2}$ at $s=0,$ we see that $E(z,s)$ is holomorphic
    at $s=0$.  Finally, since $\T(z,s)$ is holomorphic at $s=1$, we see that $E(z,s)$ has a simple pole at $s=1$ and that
    $$R_1(\T(s)E(z,s))=\frac{1}{2}.$$

    Observe that $W_{s}=W_{1-s}$ since the Bessel function $K_v$ is an even function of $v$ and $W_s=K_{s-1/2}$.  Moreover, it is clear that the
    divisor sum
    $$\sum_{ab=|r|}\left(\frac{a}{b}\right)^{s-1/2}$$ is invariant under $s\mapsto 1-s$.  We thus have
    \begin{align*}
        \T(1-s)E(z,1-s)&=\T(1-s)y^{1-s}+\T(1/2-s)y^{s}+\sum_{r\neq 0}\frac{W_{1-s}(rz)}{\sqrt{|r|}}\left(\sum_{ab=|r|}\left(\frac{a}{b}\right)^{1/2-s}\right)\\
        &=\T(1-s)y^{1-s}+\T(1/2-s)y^{s}+\sum_{r\neq 0}\frac{W_{s}(rz)}{\sqrt{|r|}}\left(\sum_{ab=|r|}\left(\frac{b}{a}\right)^{s-1/2}\right).
    \end{align*}
    Since $\T(s)=\pi^{-s}\G(s)\z(2s),$ we have $\T(1/2-s)=\T(s)$.  It follows that
    $$\T(1-s)E(z,1-s)=\T(s-1/2)y^{1-s}+\T(s)y^{s}+\sum_{r\neq 0}\frac{W_{s}(rz)}{\sqrt{|r|}}\left(\sum_{ab=|r|}\left(\frac{b}{a}\right)^{s-1/2}\right)=\T(s)E(z,s).$$
\end{exer}

\begin{exer}
    Let $f$ be a normalized Hecke Eigenform with fourier expansion
    $$\sum_{n=1}^{\infty} a_f(n)n^{\frac{k-1}{2}}q^n.$$  For a prime $p$, write $a_f(p)=\a+\b$ with $\a\b=1$.  Recall that we have the
    relations
    $$a_f(m)a_f(n)=\sum_{d|(m,n)}a_f\left(\frac{mn}{d^2}\right)$$ and
    $$a_f(p^n)=\a^n+\a^{n-1}\b+\cdots+\b^n=\frac{\a^{n+1}-\b^{n+1}}{\a-\b},$$ with appropriate interpretation if $\a=\b$.  We can therefore write
    \begin{align*}
        \z(2s)\sum_{n=1}^{\infty}\frac{a_f(n)\bar{a}_f(n)}{n^s}&=\prod_{p}\frac{1}{1-p^{-2s}}\sum_{n=0}^{\infty}\frac{a_f(p^n)\bar{a}_f(p^n)}{p^{ns}}\\
        &=\prod_{p}\frac{(1-p^{-2s})^{-1}}{(\a-\b)(\ba-\bb)}\sum_{n=0}^{\infty}\frac{(\a^{n+1}-\b^{n+1})(\ba^{n+1}-\bb^{n+1})}{p^{ns}}\\
                &=\prod_{p}\frac{(1-p^{-2s})^{-1}}{(\a-\b)(\ba-\bb)}\left(\frac{\a\ba}{1-\frac{\a\ba}{p^s}}-\frac{\a\bb}{1-\frac{\a\bb}{p^s}}
                -\frac{\b\ba}{1-\frac{\b\ba}{p^s}}+\frac{\b\bb}{1-\frac{\b\bb}{p^s}}\right)\\
       &=\prod_{p}\frac{(1-p^{-2s})^{-1}}{(\a-\b)(\ba-\bb)}\left(\frac{1}{\b\bb-\frac{1}{p^s}}-\frac{1}{\b\ba-\frac{1}{p^s}}
                -\frac{1}{\a\bb-\frac{1}{p^s}}+\frac{1}{\a\ba-\frac{1}{p^s}}\right)\\
&=\prod_{p}\frac{(1-p^{-2s})^{-1}}{(\a-\b)(\ba-\bb)}\left(\frac{\a\ba+\b\bb-\frac{2}{p^s}}{1-\frac{1}{p^s}(\a\ba+\b\bb)+\frac{1}{p^{2s}}}-
\frac{\a\bb+\ba\b-\frac{2}{p^s}}{1-\frac{1}{p^s}(\a\bb+\b\ba)+\frac{1}{p^{2s}}}\right)\\
&=\prod_{p}\left(\frac{1}
{\left(1+\frac{1}{p^{2s}}\right)^2-\frac{1}{p^s}\left(1+\frac{1}{p^{2s}}\right)(\a+\b)(\ba+\bb)+\frac{1}{p^{2s}}(\a^2+\b^2+\ba^2+\bb^2)}\right)\\
    \end{align*}
    Since we have normalized $f$ so that its fourier coefficients are precisely the eigenvalues of the $T_n$, and since these eigenvalues
    are real, it follows that
    $$\a+\b=\ba+\bb$$ and
    $$\a^2+\b^2=(\a+\b)^2-2=(\ba+\bb)^2-2=\ba^2+\bb^2.$$
    Therefore, we have
    \begin{align*}
        \z(2s)\sum_{n=1}^{\infty}\frac{a_f(n)\bar{a}_f(n)}{n^s}
&=\prod_{p}\left(\frac{1}
{\left(1+\frac{1}{p^{2s}}\right)^2-\frac{1}{p^s}\left(1+\frac{1}{p^{2s}}\right)(\a+\b)^2+\frac{2}{p^{2s}}(\a^2+\b^2)}\right)\\
&=\prod_{p}\left(\frac{1}
{\left(1+\frac{1}{p^{2s}}\right)^2-\left(1+\frac{1}{p^{2s}}\right)\frac{\a^2+\b^2}{p^s}-\frac{2}{p^s}\left(1+\frac{1}{p^{2s}}\right)
+\frac{2}{p^{s}}\frac{\a^2+\b^2}{p^s}}\right)\\
&=\prod_{p}\left(\frac{1}
{\left(1-\frac{\a^2+\b^2}{p^s}+\frac{1}{p^{2s}}\right)\left(1-\frac{2}{p^s}+\frac{1}{p^{2s}}\right)}\right)\\
&=\prod_{p}\left(1-\frac{\a^2}{p^s}\right)^{-1}\left(1-\frac{\b^2}{p^s}\right)^{-1}\left(1-\frac{1}{p^s}\right)^{-2},
    \end{align*}
as required.
\end{exer}

\begin{exer}
    Let $\F$ be the standard fundamental domain for $\SL{\Z}/\H$ and let $\{\gamma_i\}$ for $1\leq i\leq n$ be a complete
    set of coset representatives for $\SL{\Z}/\G$ where $\G\subset \SL{\Z}$ is a subgroup of finite index.
    Then we claim that no two interior points of
    $$D=\bigcup_{i=1}^n \gamma_i \F$$ are $\G$ equivalent.  For suppose to the contrary that $z,gz\in D$ for some $g\in \G$.
    If $z,gz\in \gamma_i\F$ for some $i$ then $\gamma_i^{-1}z,\gamma_i^{-1}gz\in \F$ are two $\SL{\Z}$ equivalent interior (see below) points of $\F$,
    so they must be the same point, implying that $z=gz$.  Otherwise, $z\in \gamma_i\F$ and $gz\in \gamma_j\F$
    for some $i\neq j$.  Then $\gamma_i^{-1}z,\gamma_j^{-1}gz\in\F$.
    Since we have supposed $z$ to be an interior point of $\gamma_i\F$, we see that $\gamma_i^{-1}z,\gamma_j^{-1}gz$ are interior
    points of $\F$ with
    $$\gamma_j^{-1}g\gamma_i(\gamma_i^{-1}z)=\gamma_j^{-1}gz.$$  Since any interior point of $\F$ has trivial stabilizer in
    $\SL{\Z}$, it follows that  $\gamma_i^{-1}z=\gamma_j^{-1}gz$ and
    $$g\gamma_i=\gamma_j.$$
    But since $\gamma_i,\gamma_j$ are distinct coset representatives of $\G$ in $\SL{\Z}$, they cannot be $\G$ equivalent.  This is
    a contradiction, and no two interior points of $D$ are $\G$ equivalent.

\end{exer}

\end{document}